Urgently need the general formula of the sequence an

help! The general formula of the sequence an

S(n 1) = 4an 2..........(A)

Sn = 4a(n-1) 2.......(B)

(A)-(B) Obtain, a(n 1) = 4an - 4a(n-1)

Transfer the terms, a(n 1) - 2an = 2an - 4a(n-1) = 2[an - a(n-1)]

Suppose bn = a(n 1) - 2an

Then, bn = 2b(n-1) q = 2

According to the question, S2 = a1 a2 = 4a1 2

Because a1 = 1 a2 = 5

So, b1 = a2 - 2a1 = 3

So, bn = b1*q^(n-1) = 3 * 2^(n-1)

That is a(n 1) - 2an = 3 * 2^(n-1)

2[an - 2a(n-1)] = 3 * 2^(n-2) * 2 = 3 * 2^(n-1)

2^2*[a(n-1) - 2a(n-2)] = 3 * 2^(n-3) * 2^2 = 3 * 2^(n-1)

: :

: :

: :

2^(n-1)*(a2 - 2a1) = 3 * 2^(n-1)

The above are n formulas, add the above formulas,

Get a(n 1) - 2^n*a1 = 3 * 2^(n-1) * n

a(n 1) - 2^n = 3 * 2^(n-1) * n

a(n 1) = 2^n 3 * 2^(n-1) * n

So an = 2^(n-1) 3 * 2^(n-2) * (n-1)

= 2^(n-2) * (3n-1)

The formula of the general term of the sequence

Is your title wrong? If a2 a4=5/4, a1 a5=1/4, a1 and q have no solution. .

a1*a5=1/4 is almost the same.

In that case, there are a2*a4=a1*a5=1/4, a2 a4=5/4

Solving the system of equations, we get: a2=1, a4=1/4 or a2=1/4, a4=1 (discarded because 0

= 4-(1/2)^(n-2)-n*(1/2)^*(n-1)

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